How do you differentiate # y =-ln [ 3+ (9+x^2) / x]#? Calculus Basic Differentiation Rules Chain Rule 1 Answer Øko Dec 24, 2017 #dy/dx=-1/(3+9/x+x)*(1-9/x^2)# Explanation: #y=-ln(3+(9+x^2)/x)=-ln(3+9/x+x)# Use the chain rule #dy/dx=dy/(du)*(du)/dx# Let #y=-ln(u)# then #dy/(du)=-1/u# And #u=3+9/x+x# then #(du)/(dx)=-9/x^2+1# #dy/dx=-1/u*(1-9/x^2)# Substitute #u=3+9/x+x# #dy/dx=-1/(3+9/x+x)*(1-9/x^2)# Answer link Related questions What is the Chain Rule for derivatives? How do you find the derivative of #y= 6cos(x^2)# ? How do you find the derivative of #y=6 cos(x^3+3)# ? How do you find the derivative of #y=e^(x^2)# ? How do you find the derivative of #y=ln(sin(x))# ? How do you find the derivative of #y=ln(e^x+3)# ? How do you find the derivative of #y=tan(5x)# ? How do you find the derivative of #y= (4x-x^2)^10# ? How do you find the derivative of #y= (x^2+3x+5)^(1/4)# ? How do you find the derivative of #y= ((1+x)/(1-x))^3# ? See all questions in Chain Rule Impact of this question 1290 views around the world You can reuse this answer Creative Commons License